$ A = \left[\begin{array}{r}-2 \\ 5 \\ 5\end{array}\right]$ $ C = \left[\begin{array}{rr}3 & 2\end{array}\right]$ What is $ A C$ ?
Explanation: Because $ A$ has dimensions $(3\times1)$ and $ C$ has dimensions $(1\times2)$ , the answer matrix will have dimensions $(3\times2)$ $ A C = \left[\begin{array}{r}{-2} \\ {5} \\ \color{gray}{5}\end{array}\right] \left[\begin{array}{rr}{3} & \color{#DF0030}{2}\end{array}\right] = \left[\begin{array}{rr}? & ? \\ ? & ? \\ ? & ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ A$ , with the corresponding elements in column $j$ of the second matrix, $ C$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ A$ with the first element in ${\text{column }1}$ of $ C$ , then multiply the second element in ${\text{row }1}$ of $ A$ with the second element in ${\text{column }1}$ of $ C$ , and so on. Add the products together. $ \left[\begin{array}{rr}{-2}\cdot{3} & ? \\ ? & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ A$ with the corresponding elements in ${\text{column }1}$ of $ C$ and add the products together. $ \left[\begin{array}{rr}{-2}\cdot{3} & ? \\ {5}\cdot{3} & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 1, column 2 of the answer matrix, multiply the elements in ${\text{row }1}$ of $ A$ with the corresponding elements in $\color{#DF0030}{\text{column }2}$ of $ C$ and add the products together. $ \left[\begin{array}{rr}{-2}\cdot{3} & {-2}\cdot\color{#DF0030}{2} \\ {5}\cdot{3} & ? \\ ? & ?\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{rr}{-2}\cdot{3} & {-2}\cdot\color{#DF0030}{2} \\ {5}\cdot{3} & {5}\cdot\color{#DF0030}{2} \\ \color{gray}{5}\cdot{3} & \color{gray}{5}\cdot\color{#DF0030}{2}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{rr}-6 & -4 \\ 15 & 10 \\ 15 & 10\end{array}\right] $